Làm các phép tính sau:

a) $\frac{y}{2x^2-xy}+\frac{4x}{y^2-2xy}$;             b) $\frac{1}{x+2}+\frac{3}{x^2-4}+\frac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}$;

c) $\frac{1}{x+2}+\frac{1}{\left(x+2\right)\left(4x+7\right)}$;          d) $\frac{1}{x+3}+\frac{1}{\left(x+3\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(4x+7\right)}.$

Giải

a) $\frac{y}{2x^2-xy}+\frac{4x}{y^2-2xy}=\frac{y}{x(2x-y)}+\frac{4x}{y\left(y-2x\right)}=\frac{y}{x(2x-y)}+\frac{-4x}{y\left(2x-y\right)}$

$=\frac{y^2}{xy(2x-y)}+\frac{-4x^2}{xy(2x-y)}=\frac{y^2-4x^2}{xy(2x-y)}$

= $=\frac{\left(y-2x\right)\left(y+2x\right)}{xy(2x-y)}=\frac{-\left(2x-y\right)\left(y+2x\right)}{xy(2x-y)}=\frac{-2x-y}{xy}$

b) $\frac{1}{x+2}+\frac{3}{x^2-4}+\frac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}=\frac{1}{x+2}+\frac{3}{\left(x-2\right)\left(x+2\right)}+\frac{x-14}{{\left(x+2\right)}^2\left(x-2\right)}$

$=\frac{\left(x-2\right)\left(x+2\right)}{{\left(x+2\right)}^2\left(x-2\right)}+\frac{3\left(x+2\right)}{{\left(x+2\right)}^2\left(x-2\right)}+\frac{x-14}{{\left(x+2\right)}^2\left(x-2\right)}=\frac{\left(x-2\right)\left(x+2\right)+3\left(x+2\right)+x-14}{{\left(x+2\right)}^2\left(x-2\right)}$

$$\begin{gathered} =\frac{x^2-4+3x+6+x-14}{{\left(x+2\right)}^2\left(x-2\right)}=\frac{x^2+4x-12}{{\left(x+2\right)}^2\left(x-2\right)} \\ =\frac{x^2-2x+6x-12}{{\left(x+2\right)}^2\left(x-2\right)}=\frac{x\left(x-2\right)+6\left(x-2\right)}{{\left(x+2\right)}^2\left(x-2\right)} \\ =\frac{\left(x-2\right)\left(x+6\right)}{{\left(x+2\right)}^2\left(x-2\right)}=\frac{x+6}{{\left(x+2\right)}^2} \end{gathered}$$

c) $$\begin{gathered} \frac{1}{x+2}+\frac{1}{\left(x+2\right)\left(4x+7\right)}=\frac{\left(4x+7\right)}{\left(x+2\right)\left(4x+7\right)}+\frac{1}{\left(x+2\right)\left(4x+7\right)} \\ =\frac{4x+7+1}{\left(x+2\right)\left(4x+7\right)}=\frac{4x+8}{\left(x+2\right)\left(4x+7\right)}=\frac{4(x+2)}{\left(x+2\right)\left(4x+7\right)}=\frac{4}{4x+7} \end{gathered}$$

d) $$\begin{gathered} \frac{1}{x+3}+\frac{1}{\left(x+3\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(4x+7\right)}=\frac{x+2}{\left(x+3\right)\left(x+2\right)}+\frac{1}{\left(x+3\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(4x+7\right)} \\ =\frac{x+3}{\left(x+3\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(4x+7\right)}=\frac{1}{x+2}+\frac{1}{\left(x+2\right)\left(4x+7\right)} \\ =\frac{4x+7}{\left(x+2\right)\left(4x+7\right)}+\frac{1}{\left(x+2\right)\left(4x+7\right)}=\frac{4x+8}{\left(x+2\right)\left(4x+7\right)}=\frac{4(x+2)}{\left(x+2\right)\left(4x+7\right)}=\frac{4}{4x+7} \end{gathered}$$